Coderforce 444cDZY Loves Colors(线段树)

C. DZY Loves Colors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of thei-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

1. Paint all the units with numbers between l and r (both inclusive) with color x.
2. Ask the sum of colorfulness of the units between l and r (both inclusive).

Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Sample test(s)

input

3 31 1 2 41 2 3 52 1 3

output

8

input

3 41 1 3 42 1 12 2 22 3 3

output

321

input

10 61 1 5 31 2 7 91 10 10 111 3 8 121 1 10 32 1 10

output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.

题目大意：n个位置，m个操作，开始第i个位置的颜色是i，然后1 a b c,把【a,b】区间都染成c，每个位置改变的值为开始的颜色x与染成的那个颜色的绝对值abs（x-z），2 a b，区间【a,b】的改变值的总和

ac代码

14649342Practice:
kxh1995444C - 26MS C++Accepted233 ms9400 KB2015-12-04 17:17:212015-12-04 17:17:22Compare

#include<stdlib.h>#include<string.h>#include<iostream>#include<algorithm>#include<stdio.h>#include<math.h>#include<limits.h>#define N 100010#define LL long longusing namespace std;struct s{    LL col,colf,flag;}node[N<<2];void build(int l,int r,int tr){    node[tr].col=node[tr].colf=node[tr].flag=0;    if(l==r)    {        node[tr].col=l;        return;    }    int mid=(l+r)>>1;    build(l,mid,tr<<1);    build(mid+1,r,tr<<1|1);}LL ABS(LL x){    if(x<0)        return -x;    return x;}void pushdown(int tr,int m){    if(node[tr].flag)    {        node[tr<<1].flag+=node[tr].flag;        node[tr<<1|1].flag+=node[tr].flag;        node[tr<<1].colf+=node[tr].flag*(m-(m>>1));        node[tr<<1|1].colf+=node[tr].flag*(m>>1);        node[tr].flag=0;    }    if(node[tr].col)    {        node[tr<<1].col=node[tr<<1|1].col=node[tr].col;    }}void pushup(int tr){    node[tr].colf=node[tr<<1].colf+node[tr<<1|1].colf;    if(node[tr<<1].col==node[tr<<1|1].col)        node[tr].col=node[tr<<1].col;    else        node[tr].col=0;}void update(int L,int R,LL val,int l,int r,int tr){    if(L<=l&&r<=R)    {        if(node[tr].col)        {            node[tr].colf+=(r-l+1)*ABS(node[tr].col-val);            node[tr].flag+=ABS(node[tr].col-val);            node[tr].col=val;            return;        }    }    pushdown(tr,r-l+1);    int mid=(l+r)>>1;    if(L<=mid)        update(L,R,val,l,mid,tr<<1);    if(R>mid)        update(L,R,val,mid+1,r,tr<<1|1);    pushup(tr);}LL query(int L,int R,int l,int r,int tr){    if(L<=l&&r<=R)    {        return node[tr].colf;    }    pushdown(tr,r-l+1);    LL a,b;    a=0;    b=0;    int mid=(l+r)>>1;    if(L<=mid)    {        a=query(L,R,l,mid,tr<<1);    }    if(R>mid)    {        b=query(L,R,mid+1,r,tr<<1|1);    }    return (a+b);}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        build(1,n,1);        while(m--)        {            int op;            scanf("%d",&op);            if(op==1)            {                int a,b;                LL c;                scanf("%d%d%lld",&a,&b,&c);                update(a,b,c,1,n,1);            }            else            {                int a,b;                scanf("%d%d",&a,&b);                printf("%lld\n",query(a,b,1,n,1));            }        }    }}